Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Main {

	static Scanner in = new Scanner(System.in);
	static int X = 0, Y = 0, minNum = 501;

	public static void main(String[] args) throws IOException {

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		int grid[][] = new int[n + 1][m + 1];
		boolean hasZeroR[] = new boolean[n + 1];
		boolean hasZeroC[] = new boolean[m + 1];
		int minR[] = new int[n + 1];
		int minC[] = new int[m + 1];
		Arrays.fill(hasZeroC, false);
		Arrays.fill(hasZeroR, false);
		Arrays.fill(minC, 0);
		Arrays.fill(minR, 0);

		for (int i = 1; i <= n; i++) {
			st = new StringTokenizer(br.readLine());
			for (int j = 1; j <= m; j++) {
				grid[i][j] = Integer.parseInt(st.nextToken());
				if (grid[i][j] == 0) {
					hasZeroC[j] = hasZeroR[i] = true;
				}
			}
		}

		for (int i = 1; i <= n; i++) {
			minR[i] = 501;
			for (int j = 1; j <= m; j++) {
				minR[i] = Math.min(minR[i], grid[i][j]);
			}
			// System.out.println("min row " + i +" : " + minR[i]);
		}
		for (int j = 1; j <= m; j++) {
			minC[j] = 501;
			for (int i = 1; i <= n; i++) {
				minC[j] = Math.min(minC[j], grid[i][j]);
			}
			// System.out.println("min col " + j +" : " + minC[j]);
		}
		StringBuffer ans = new StringBuffer();
		int moves = 0;
		if(n<=m)
		{
			for (int i = 1; i <= n; i++) {
				if (hasZeroR[i])
					continue;

				for (int k = 0; k < minR[i]; k++) {
					ans.append("row " + i + "\n");
					moves++;
				}

				for (int j = 1; j <= m; j++) {
					grid[i][j] -= minR[i];
					minC[j] = Math.min(minC[j], grid[i][j]);
					if (minC[j] == 0)
						hasZeroC[j] = true;
				}

			}

			for (int j = 1; j <= m; j++) {
				if (hasZeroC[j])
					continue;

				for (int k = 0; k < minC[j]; k++) {
					ans.append("col " + j + "\n");
					moves++;
				}
				for (int i = 1; i <= n; i++) {
					grid[i][j] -= minC[j];
					minR[i] = Math.min(minR[i], grid[i][j]);
				}

			}
		}else {
			for (int j = 1; j <= m; j++) {
				if (hasZeroC[j])
					continue;

				for (int k = 0; k < minC[j]; k++) {
					ans.append("col " + j + "\n");
					moves++;
				}
				for (int i = 1; i <= n; i++) {
					grid[i][j] -= minC[j];
					minR[i] = Math.min(minR[i], grid[i][j]);
				}

			}
			for (int i = 1; i <= n; i++) {
				if (hasZeroR[i])
					continue;

				for (int k = 0; k < minR[i]; k++) {
					ans.append("row " + i + "\n");
					moves++;
				}

				for (int j = 1; j <= m; j++) {
					grid[i][j] -= minR[i];
					minC[j] = Math.min(minC[j], grid[i][j]);
					if (minC[j] == 0)
						hasZeroC[j] = true;
				}

			}
		}

		boolean fail = false;

		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++)
				if (grid[i][j] > 0) {
					fail = true;
					break;
				}
		if (fail) {
			System.out.println(-1);
			return;
		}
		System.out.println(moves);
		System.out.println(ans);
	}
}